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Current Question (ID: 20539)

Question:

$\text{An alternating voltage } V(t) = 220 \sin(100\pi t) \text{ volts is applied to a purely resistive load of } 50 \, \Omega. \text{ The time taken for the current to rise from half of the peak value to the peak value is:}$ $1. \, 5 \, \text{ms}$ $2. \, 7.2 \, \text{ms}$ $3. \, 3.3 \, \text{ms}$ $4. \, 2.2 \, \text{ms}$

Options:

1. $5 \, \text{ms}$
2. $7.2 \, \text{ms}$
3. $3.3 \, \text{ms}$
4. $2.2 \, \text{ms}$

Solution:

$\text{Hint: } t = \frac{\theta}{\omega}$ $\text{Step: Find the time taken for the current to rise from half of the peak value to the peak value.}$ $\text{From the question, we can have the value of alternating voltage;} \Rightarrow V(t) = 220 \sin(100\pi t)$ $\text{Now, the current value is given by;} \Rightarrow I(t) = \frac{V(t)}{R}$ $\text{Substitute the known values we get;} \Rightarrow I(t) = \frac{220}{50} \sin(100\pi t)$ $\Rightarrow I = I_m \sin(\omega t) \text{ Now, the time is given by the formula;} \Rightarrow t = \frac{\theta}{\omega} \text{ The sine waveform is maximum at the angle } \frac{\pi}{2}.$ $t_1 = \frac{\pi}{2} \times \frac{1}{100\pi} = \frac{1}{200} \text{ sec}$ $\text{For, } I = \frac{I_m}{2}$ $\Rightarrow \frac{I_m}{2} = I_m \sin(100\pi t_2)$ $\Rightarrow \sin \frac{\pi}{6} = \sin(100\pi t_2) \left[ \therefore \sin \frac{\pi}{6} = \sin 60^\circ = \frac{1}{2} \right]$ $\Rightarrow t_2 = \frac{1}{600} \text{ s}$ $\text{Now, } t_{\text{req}} = \frac{1}{200} - \frac{1}{600}$ $\Rightarrow t_{\text{req}} = \frac{2}{600} = \frac{1}{300} \text{ s} = 3.3 \text{ ms}$ $\text{Hence, option (3) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}